The decomposition reaction involved 7.8 grams of AlOH3 What masses of oxide and water were formed?

1. We write down the reaction of decomposition of aluminum hydroxide:

2Al (OH) 3 = Al2O3 + 3H2O;

2.find the chemical amount of hydroxide:

n (Al (OH) 3) = m (Al (OH) 3): M (Al (OH) 3);

M (Al (OH) 3) = 27 + 3 * 16 + 3 * 1 = 78 g / mol;

n (Al (OH) 3) = 7.8: 78 = 0.1 mol;

3. find the amounts of aluminum oxide and water:

n (Al2O3) = n (Al (OH) 3): 2 = 0.05 mol;

n (H2O) = n (Al2O3) = 0.05 mol;

4.Calculate the masses of oxide and water:

m (Al2O3) = n (Al2O3) * M (Al2O3) = 0.05 * 102 = 5.1 g;

m (H2O) = n (H2O) * M (H2O) = 0.05 * 18 = 0.9 g.

Answer: 5.1 g Al2O3, 0.9 g H2O.